Question: Divide the following complex numbers. $ \dfrac{-8-14i}{4+2i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4-2i}$ $ \dfrac{-8-14i}{4+2i} = \dfrac{-8-14i}{4+2i} \cdot \dfrac{{4-2i}}{{4-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-14i) \cdot (4-2i)} {(4+2i) \cdot (4-2i)} = \dfrac{(-8-14i) \cdot (4-2i)} {4^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-14i) \cdot (4-2i)} {(4)^2 - (2i)^2} = $ $ \dfrac{(-8-14i) \cdot (4-2i)} {16 + 4} = $ $ \dfrac{(-8-14i) \cdot (4-2i)} {20} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-14i}) \cdot ({4-2i})} {20} = $ $ \dfrac{{-8} \cdot {4} + {-14} \cdot {4 i} + {-8} \cdot {-2 i} + {-14} \cdot {-2 i^2}} {20} $ Evaluate each product of two numbers. $ \dfrac{-32 - 56i + 16i + 28 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{-32 - 56i + 16i - 28} {20} = \dfrac{-60 - 40i} {20} = -3-2i $